Correlation between electronic band structure of an atomic species

[thinktank1985]

Hii,

I was wondering whether there is any correlation between the energy eigenvalues of a single atom (say He) with the electronic band structure of a collection of He atoms.

for example, for He atoms the lowest energy would be the ground state, with 2 electrons. So will the lowest band for a collection of He atoms be centered about the ground state of the He atom? Do the other bands in the electronic band structure of He, follow a similar pattern, in which they are situation about certain eigenvalues of the He atom?

 

[cgk]

The band structure is essentially a representation of the energy spectrum of the orbital eigenvalues of the periodic system (NOT of the many-body electronic states!). That is, it is not real, it is intrinsically bound to a one-particle mean field approximation (say, Kohn-Sham or Hartree-Fock).

These orbital energies depend on the binding situation. Orbitals which are not involved in binding to any other atoms (say, core orbitals) are more or less unchanged also in the solid, as are their orbital eigenvalues (appar from some constant shift for all orbitals in common). That means, if you look into the 1s bands of a helium crystal, for example, you will see next to no disperson and essentially flat bands --- because the crystal splitting (due to the chemical environment!) is very small. In the helium crystal the same actually also happens for the 2s band, because noble gas crystals are not bound by covalent bonds, but by undirected dispersive interactions (and latent heat stored in zero point vibrations; these apparently account for up to 30% in the cohesive energy of such crystals: http://prb.aps.org/abstract/PRB/v62/i9/p5482_1)

If, on the other hand, the orbitals corresponding to the bands *are* involved in binding, then you can generally not say much about the relationship between the atomic orbital levels and the crystal orbital levels. The energies will typically be in the same order of magnitude (mind the constant shift), but that is all.